An Investigation into the effect of surface area on the activity of catalase in potato.

Authors Avatar

An Investigation into the effect of surface area on the activity of catalase in potato

An experiment to find out how the application of different surface areas of the same sized potato chip, effects the reaction rate of catalase in the potato chip, whilst submerged in hydrogen peroxide at a controlled temperature and pH level.   

Introduction:

Enzymes are large proteins and catalysts and increase the speed of a chemical reaction without undergoing any permanent chemical change. They are neither used up in the reaction nor do they appear as reaction products. 2  Enzymes, such as Catalase, are globular protein molecules which are found in living cells. 3 In their globular structure, one or more polypeptide chains twist and fold, bringing together a small number of amino acids to form the active site, or the location on the enzyme where the substrate binds and the reaction takes place. Enzyme and substrate fail to bind if their shapes do not match exactly. This ensures that the enzyme does not participate in the wrong reaction, as they are all very specific, as each enzyme just performs one particular reaction. 4

Catalase is an enzyme found in food such as potato and liver. It is used for removing Hydrogen Peroxide from the cells. Hydrogen Peroxide is the poisonous by-product of the affects of metabolism. Catalase speeds up the decomposition of Hydrogen Peroxide into the two products,  water and oxygen, as shown in the equations below.

       

 Catalase

HYDROGEN PEROXIDE                      WATER   +  OXYGEN

 Catalase

                   2H2O2                           2H2O2   +  O2

It is able to speed up the decomposition of Hydrogen Peroxide because the shape of its active site matches the shape of the Hydrogen Peroxide molecule. This type of reaction where a molecule is broken down into smaller pieces is called an anabolic reaction.

When the catalase comes into contact with hydrogen peroxide, it turns the H2O2 into water (H2O) and oxygen gas (O2), as can be seen in the formula displayed above. Which when these products have been released, the enzyme is then ready to bind with a new substrate. Catalase does this extremely efficiently – with up to one molecule of catalase breaking 40 million molecules of hydrogen peroxide each second. 5  The bubbles that can be seen in the foam on the surface of the hydrogen peroxide, are pure oxygen bubbles being created by the catalase, in the binding of the enzyme molecules and the substrate molecules through the active site. 1 (As shown on next page).

4

The activity of catalase is strongly affected by changes in pH (left graph shown below) and temperature (right graph shown below). Each enzyme works best at a certain pH and temperature, its activity decreasing at values above and below that point, known as its optimum pH or temperature. This is not surprising considering the importance of tertiary structure (i.e. shape) in enzyme function and non-covalent forces, e.g., ionic interactions and hydrogen bonds, in determining that shape. 5  

It is known that the enzyme catalase works best at pH 7, but is stable between pH 6 to 8, and has an optimum temperature of around 40 °C, whereas it is stable between 0 and 50 °C.

        Through increase in surface area of the potato chip, it is proven that the rate of reaction will increase as there is more catalase in contact with the hydrogen peroxide, and so more collisions between the substrate and enzyme molecules will result. I have below calculated the exact size in millimeters-2 of potato chips that will be used within my experiment. My reasons for using millimeters within the calculations of experiment, consist of solely upon the accuracy of the experiment overall, as I did not want to make this a limiting factor by choosing a larger scale, such as centimeters. I have used the following formula shown on the next page in order to calculate my surface areas within the experiment;

SURFACE AREA OF A CYLINDER = (2πrh + 2πr –2) x n

Where, r = radius of cylinder

          h = height of cylinder

          n = number of potato chips

  1. Where, r = 4

h = 40

                n = 1

        Therefore,

SURFACE AREA = (2πrh + 2πr –2) x n

       = (2π x 4 x 40 + 2π x 4–2) x 1

                                                       = (1005.309649 + 100.5309649) x 1

                                                       = 1105.841mm–2  (to 3 decimal places)

  1. Where, r =  4

h = 20

n = 2

        Therefore,

SURFACE AREA = (2πrh + 2πr –2) x n

       = (2π x 4 x 20 + 2π x 4–2) x 2

                                                       = (502.6548246 + 100.5309649) x 2

                                                       = 1206.372mm–2  (to 3 decimal places)

  1. Where, r = 4

h = 10

n = 4

        Therefore,

SURFACE AREA = (2πrh + 2πr –2) x n

       = (2π x 4 x 10 + 2π x 4–2) x 4

                                                       = (251.3274123 + 100.5309649) x 4

                                                       = 1407.434mm–2  (to 3 decimal places)

  1. Where, r = 4

h = 5

n = 8

SURFACE AREA = (2πrh + 2πr –2) x n

       = (2π x 4 x 5 + 2π x 4–2) x 8

                                                       = (125.6637061 + 100.5309649) x 8

                                                       = 1809.557mm–2  (to 3 decimal places)

Prediction:

Through research within my introduction, I have discovered that catalase has an optimum temperature of 40 °C, where it is stable between 0 and 50 °C. This is why I have chosen to run my experiment at the temperature of 38 °C, so that it will never be completely at its optimum, as this may risk the chance of the enzyme being denatured if the temperature were to go slightly above this. As although it would be still stable up to 50 °C, the rate of reaction would begin to slow due to some of the enzyme molecules fitting less well into the active site of the catalase enzyme. With eventually the hydrogen peroxide substrate molecule no-longer fitting at all, or not being able to be held in its corrects position for the reaction to occur. Which is known as the enzyme being denatured, which is often irreversible.

In reference to the surface area of potato chips I use, I predict that when I increase the surface area of the potato chip, the reaction rate of catalase reacting with the hydrogen peroxide will increase. Thus, causing more oxygen to be produced and released from the potato chip, so that the collection of oxygen in the glass syringe will be much greater. This is because as the surface area of the potato chip increases, there is more of the enzyme catalase being exposed to the substrate molecule of hydrogen peroxide. Meaning that there are more enzyme molecules available to bind with the substrate molecules of hydrogen peroxide; causing the reaction to speed up. I also predict that at the beginning of the experiment, the ‘Initial rate of reaction’, the reaction rate of catalase and hydrogen peroxide will be much faster. This is due to there being more substrate molecules available at this time to react with the enzyme molecules active site, causing the release of the two products; water and oxygen. I believe that after this initial rate of reaction, the reaction will slow, due to there being less substrate molecules to bind with the catalase enzyme molecules, and release the two products. This I believe will be clearly visible when I collate the results into graphical terms. Showing of the volume of oxygen collected in the glass syringe, at 30-second intervals of the experiment, to slow, causing the curve I envisage to be produced, to gently level off, reducing in its gradient.

Join now!

Preliminary work:

For my real experiment I have completed a pilot study, which has enabled me to test my choices of apparatus and methods that I will use, so that any alterations can be made in the event of there being a problem.

        During this preliminary study I have encountered a number of consequential problems. One of these involved the container in which to hold my experiment within, being that of a conical flask. However, through the pilot study I found that the size and volume of the conical flask affected the correct working of the glass syringe. It ...

This is a preview of the whole essay